Limits Misconception Differentiation

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lhh2003
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What I gathered was that the limit is always 0 because it gives the most accurate (the exact so to speak) value of the gradient of the tangent of a point on a curve.

What confuses me, however, is how the derivative formula works if h = 0 :

Let y = f(x)

Point A ( x0 , f(x0) )

Point B (x0 + h , f(x0 + h) )

Let x0 = 5 and h = 0 , with lim h--> 0

f ' (5) = ( f ( 5 + 0) - f( 5 ) ) / 0


^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.

Thank you.
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hustlr
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thats why you have to insert whatever function your going to use first and THEN insert h=0 because you cant simply just put 0 in at the start as its impossible

for example let f(x)= x²-1, insert this into your general gradient function and THEN insert h=0
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mqb2766
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(Original post by lhh2003)
What I gathered was that the limit is always 0 because it gives the most accurate (the exact so to speak) value of the gradient of the tangent of a point on a curve.

What confuses me, however, is how the derivative formula works if h = 0 :

Let y = f(x)

Point A ( x0 , f(x0) )

Point B (x0 + h , f(x0 + h) )

Let x0 = 5 and h = 0 , with lim h--> 0

f ' (5) = ( f ( 5 + 0) - f( 5 ) ) / 0


^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.

Thank you.
Just try it with
f(x) = 2x
What happens? The answer should be 2.
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RichE
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(Original post by lhh2003)
^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.
For polynomials you can rearrange and set h=0, but for more complicated functions like sin(x) - which differentiates to cos(x) - you do need some notion of limit. You can't just rearrange and set h=0. This tooks mathematicians generations to get completely right, so don't be fazed that you're finding it a little difficult to start.

You're already somewhat used to the idea of limits. We write pi = 3.14159265... and the ... means the expansion goes on forever and in some sense they're only equal when we have written all the decimal places. But you need a notion of limit to make clear just what that means.
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Elosant
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If you want to understand limits more rigorously, perhaps take a look at the epsilon-delta definition of limits? There is no division by zero here, and when you are taking a limit you aren't just "setting it to 0". Some functions may not even have lim f(a) = f(x), a->x for all x (these functions we call discontinuous at x), and gives rise to the notion of left and right limits if a "jump discontinuity" exists. f(x) does not even need to be defined for an arbitrary x, but you can still compute a limit for it.

https://en.wikipedia.org/wiki/Epsilon-delta
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lhh2003
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(Original post by hustlr)
thats why you have to insert whatever function your going to use first and THEN insert h=0 because you cant simply just put 0 in at the start as its impossible

for example let f(x)= x²-1, insert this into your general gradient function and THEN insert h=0
I don't understand how this method works ; if you're saying that h is 0 , then surely you are not establishing a difference between point A and point B , therefore surely you won't have two separate points to calculate a gradient with ?
Last edited by lhh2003; 10 months ago
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_gcx
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(Original post by lhh2003)
What I gathered was that the limit is always 0 because it gives the most accurate (the exact so to speak) value of the gradient of the tangent of a point on a curve.

What confuses me, however, is how the derivative formula works if h = 0 :

Let y = f(x)

Point A ( x0 , f(x0) )

Point B (x0 + h , f(x0 + h) )

Let x0 = 5 and h = 0 , with lim h--> 0

f ' (5) = ( f ( 5 + 0) - f( 5 ) ) / 0


^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.

Thank you.
This is imo the issue with limits not being introduced properly at A-level. Limits don't simply involve substituting h = 0.

What \lim_{h \to 0} f(h) = y means is that as you pick values of h closer and closer to 0 and substitute this into f, you will get values closer to closer to y.

In fact, you should be able to make f(h) as close to y as you want by picking a suitably close h. For example, you can find a h such that f(h) is within \dfrac 1 {100} of y. Try this with, for example, f(h) = h.

This edges towards a more formal definition of a limit which you don't quite need to understand at this level.

You can only necessarily say:

\displaystyle \lim_{h \to 0} \frac {f(h)} {g(h)} = \frac {f(0)} {g(0)}

if a) both f and g are continuous (they are in this case, you don't have to worry about this) and b) g(0) \ne 0. But in this case, h = 0 on the denominator, so we can't just plug in!

You want to start by simplifying the quotient:

\displaystyle \frac {f(x+h) - f(x)} h

(for example) as much as you can (within these simplifications we take h \ne 0), and then taking a limit. If you get something that's continuous at h = 0, now you can just plug it in.

For example, if we have a function f with f(x) = x^2 for all x \in \mathbb R we have for h \ne 0:

\displaystyle \frac {f(x+h) - f(x)} h = \frac {(x + h)^2 - x^2} h = \frac {2xh + h^2} h = 2x + h

There is no issue here (no division by zeroes caused, etc.) with now plugging in h = 0 to get:

\displaystyle \lim_{h \to 0} (2x + h) = 2x + 0 = 2x

giving:

f'(x) = \lim_{h \to 0} \displaystyle \frac {f(x+h) - f(x)} h = 2x

Note that simplifying it to the point that we can just "plug h = 0" in isn't possible, sometimes it's more complicated. In this case we would aim to simplify the limit so we can apply other limits we already know.

Writing \lim throughout is poor notation but perfectly acceptable at this level.

This is a fairly complicated topic that confuses many undergrads so don't worry that you're struggling.
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lhh2003
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(Original post by _gcx)
This is imo the issue with limits not being introduced properly at A-level. Limits don't simply involve substituting h = 0.

What \lim_{h \to 0} f(h) = y means is that as you pick values of h closer and closer to 0 and substitute this into f, you will get values closer to closer to y.

In fact, you should be able to make f(h) as close to y as you want by picking a suitably close h. For example, you can find a h such that f(h) is within \dfrac 1 {100} of y. Try this with, for example, f(h) = h.

This edges towards a more formal definition of a limit which you don't quite need to understand at this level.

You can only necessarily say:

\displaystyle \lim_{h \to 0} \frac {f(h)} {g(h)} = \frac {f(0)} {g(0)}

if a) both f and g are continuous (they are in this case, you don't have to worry about this) and b) g(0) \ne 0. But in this case, h = 0 on the denominator, so we can't just plug in!

You want to start by simplifying the quotient:

\displaystyle \frac {f(x+h) - f(x)} h

(for example) as much as you can (within these simplifications we take h \ne 0), and then taking a limit. If you get something that's continuous at h = 0, now you can just plug it in.

For example, if we have a function f with f(x) = x^2 for all x \in \mathbb R we have for h \ne 0:

\displaystyle \frac {f(x+h) - f(x)} h = \frac {(x + h)^2 - x^2} h = \frac {2xh + h^2} h = 2x + h

There is no issue here (no division by zeroes caused, etc.) with now plugging in h = 0 to get:

\displaystyle \lim_{h \to 0} (2x + h) = 2x + 0 = 2x

giving:

f'(x) = \lim_{h \to 0} \displaystyle \frac {f(x+h) - f(x)} h = 2x

Note that simplifying it to the point that we can just "plug h = 0" in isn't possible, sometimes it's more complicated. In this case we would aim to simplify the limit so we can apply other limits we already know.

Writing \lim throughout is poor notation but perfectly acceptable at this level.

This is a fairly complicated topic that confuses many undergrads so don't worry that you're struggling.
I can understand that we want the smallest value possible for h.

But surely saying that h = 0 is wrong, because it means that A and B become the same point, so there is no gradient to calculate (as you need to distinct points to calculate a gradient from) ?
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RichE
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(Original post by lhh2003)
I can understand that we want the smallest value possible for h.

But surely saying that h = 0 is wrong, because it means that A and B become the same point, so there is no gradient to calculate (as you need to distinct points to calculate a gradient from) ?
But there isn't a smallest value for h. In this case you'd like h to be the smallest positive real number (or some such) and that just doesn't exist. If you really want to do this properly you'll need to read up on a little analysis.

By way of a simple example, to show how you might rigorously show that sinx differentiates to cosx, you'd have to do something like the following.

(sin(x+h)-sin(x))/h = (sin(x)cos(h) + cos(x)sin(h) - sin(x))/h = sin(x)(cos(h)-1)/h + cos(x) (sin(h)/h)

You would then need to show that (cos(h)-1)/h converges to 0 as h converges to 0, and sin(h)/h converges to 1 as h converges to 0. You may know how to do these already via some geometric argument. It's clear you can't just set h=0 to show these, but if you try small values on your calculator (in radians mode) you'll see this pattern emerging.
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hustlr
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(Original post by lhh2003)
I don't understand how this method works ; if you're saying that h is 0 , then surely you are not establishing a difference between point A and point B , therefore surely you won't have two separate points to calculate a gradient with ?
im ngl i had the same trouble when i was first introduced to differentiation as well a few weeks ago

so basically to calculate gradient you need a respect to TWO points right, but if your trying to find the gradient at a specific point its different because it has a respect to one point only.

so what we do is treat two points (A and C) on either side of the point of interest (B) with a small distance apart.

reducing this distance will bring the distance closer to 0, and therefore make the true gradient of point B clearer, however it will never truly reach 0 because of the idea of the distance becoming smaller and smaller forever (infintissimaly small)

therefore all though h would never really equal 0, we can observe that it approaches 0 which is where the limit>0 comes from, using this limit illustrates the gradient with respect to two points (because there is a distance, and any distance requires two points)

ngl thats what i know personally, im certain that both of us will understand differentiation more as time goes on because thats what usually happens, we dont understand new concepts sometimes so we just need a bit of time to adjust
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